# How do you balance Al + CuCl_2 -> AlCl_3 + Cu?

Apr 13, 2018

Well this is a redox reaction....

#### Explanation:

Aluminum is oxidized to $A {l}^{3 +}$.

$A l \left(s\right) \rightarrow A {l}^{3 +} + 3 {e}^{-}$ $\left(i\right)$

And cupric ion is reduced to zerovalent copper....

$C {u}^{2 +} + 2 {e}^{-} \rightarrow C u \left(s\right)$ $\left(i i\right)$

We cross multiply to remove the electrons....$2 \times \left(i\right) + 3 \times \left(i i\right)$

$2 A l \left(s\right) + 3 C {u}^{2 +} + 6 {e}^{-} \rightarrow 3 C u \left(s\right) + 2 A {l}^{3 +} + 6 {e}^{-}$

...to give finally...

$2 A l \left(s\right) + 3 C {u}^{2 +} \rightarrow 3 C u \left(s\right) + 2 A {l}^{3 +}$

This reaction is spontaneous as written, given that aluminum is higher on the activity table.....

And so see what happens when you add some aluminum foil to a solution of copper nitrate? The foil will disintergrate....